Simpler proof that $y^3[d^2y/dx^2]$ is a constant if $y^2=ax^2+bx+c$?

Question

here's my question If $y^2=ax^2+bx+c$ then prove that $y^3[d^2y/dx^2]$ is a constant . I have solved this using the conventional method, taking square root, differentiating w.r.t to x and using chain and quotient rule But can't think of some alternative smaller and more efficient method ?? Can you ?

Also can someone suggest me any alternative to quotient rule I need something more non conventional, time saving and vastly Applicable method .

Answer 1

Differentiating both sides once, we have

$$ 2y\frac{\mathrm{d}y}{\mathrm{d}x}=2ax+b, $$

and differentiating twice, we reach

$$ 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+2y\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=2a. $$

Substituting, we find

$$ \left(\frac{2ax+b}{2y}\right)^2+y\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=a $$

$$ (2ax+b)^2+4y^3\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=4ay^2 $$

$$ 4y^3\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=4a(ax^2+bx+c)-(2ax+b)^2 $$

$$ 4y^3\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=(4a^2x^2+4abx+4ac)-(4a^2x^2+4abx+b^2) $$

$$ y^3\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=\frac{4ac-b^2}{4}. $$

We still use the chain rule but not the quotient rule.

Answer 2

Using implicit derivation: $$ (y^2)^\prime = 2yy^\prime = 2ax+b \Rightarrow y^\prime=\frac{2ax+b}{2y} $$ $$ (y^2)^{\prime\prime} = 2(y^\prime)^2+2yy^{\prime\prime} = 2a \Rightarrow y^{\prime\prime}=\frac{a-(y^\prime)^2}{y} $$ Thus \begin{align} y^3y^{\prime\prime}=y^2(a-(y^\prime)^2) &= y^2(a-\left(\frac{2ax+b}{2y}\right)^2) \\ &= ay^2-\frac{1}{4}(2ax+b)^2 \\ &= (a^2x^2+abx+ac)-\frac{1}{4}(4a^2x^2+4abx+b^2) \\ &= (a^2x^2+abx+ac)-(a^2x^2+abx+\frac{b^2}{4}) \\ &= ac - \frac{b^2}{4} \ \ \ \text{(which is constant)} \end{align}

Answer 3

Using differentials, it is easy to see that $$ 2y dy = (2ax+ b) dx \tag{1} $$ from which we can write $$ u = \frac{dy}{dx} = \frac{2ax+b}{2y} $$

Using again differentials and product rule, \begin{eqnarray} du &=& \frac{2a}{2y} dx - \frac{2ax+b}{2y^2} dy = \left[ \frac{a}{y} - \frac{(2ax+b)^2}{4y^3} \right] dx \end{eqnarray} where we have used (1).

Finally \begin{eqnarray} y^3\frac{du}{dx} = y^3\frac{d^2y}{dx^2} &=& \left[ ay^2 - \frac{(2ax+b)^2}{4} \right] = \left[ a(ax^2+bx+c) - \frac{(2ax+b)^2}{4} \right] = ac-\frac{b^2}{4} \end{eqnarray}

Answer 4

Here is a proof that uses only the fact that $u = ax^2 + bx + c$ is quadratic and hence $u''' = 0$.

We know that $y^2 = u$ and hence $2 y y' = u' , \, y^4 = u^2$. Differentiate the second identity twice: $$ 4 y^3y' = 2 u u', \quad 4 y^3 y'' + 12 y^2 (y')^2 = 2uu'' + 2 (u')^2 \, . $$ Since $4y^2(y')^2 = (u')^2$, this implies $$ 4 y^3 y'' = 2uu'' + 2 (u')^2 - 12 y^2 (y')^2 = 2 uu'' - (u')^2 \, . $$ Differentiate this one more time and use the assumption $u''' = 0$ to obtain $$ \frac{d}{dx} \left( 4 y^3 \frac{d^2y}{dx^2} \right) = 2 u u''' + 2 u' u''- 2u' u'' = 0 \, . $$ Hence $y^3 \frac{d^2y}{dx^2} $ is constant.